Flirting Sequences, Convergence (Real Analysis) Can anyone provide some insight or help into the following problem?
Definition: In a metric space, a sequence ${p_n}$ flirts with $p$ iff for each $\epsilon > 0$ there are $m,n \in \mathbb{N}$ where $m>n$ such that $0<d(p_n, p) < \epsilon$ and $d(p_n,p) < d(p_m,p)$.
Problem: If a sequence ${p_n}$ flirts with each point in a sequence ${q_n}$ and $q_n$ converges to $q$, does ${p_n}$ flirt with $q$? Prove it.
I wrote out the definitions for "${p_n}$ flirts with each point in a sequence ${q_n}$" and for "$q_n$ converges to $q$. I also wrote out the definition for "${p_n}$ flirt with $q$". I've been staring at them, trying to manipulate the first two so that they work for the third definition, but am stuck. I am not even sure how to proceed. Any help in trying to solve this problem will be helpful!
For reference, you can check this post for examples of flirting sequences.
By the way... I think the answer is no, but its more of intuition than anything.
 A: If there is an $n$ such that $q_n = q$, then it immediately follows that $(p_n)$ flirts with $q$.
Thus we need only care about the case $q_n \neq q$ for all $n$.
Let $\epsilon > 0$ be given. Choose an arbitrary $k$. Let $\delta = \frac{1}{3}d(q_k,q)$ and $\eta = \min \{\epsilon/2, \delta\}$. Since $q_n \to q$ there is an $r$ with $d(q_r,q) < \eta$. Since $(p_n)$ flirts with $q_r$ there is an $n$ with $0 < d(p_n,q_r) < d(q_r,q) < \eta$, and consequently $0 < d(p_n,q) < 2\eta < \epsilon$. It now suffices to show that there is an $m > n$ with $d(p_m,q_k) < \eta$, for then
$$d(p_m,q) \geqslant d(q_k,q) - d(p_m,q_k) > d(q_k,q) - \eta = 3\delta - \eta \geqslant 2\eta > d(p_n,q)$$
shows that $(p_n)$ flirts with $q$.
But there are only finitely many indices $\leqslant n$, thus
$$\lambda = \min \:(\{ d(p_s,q_k) : s \leqslant n, p_s \neq q_k\} \cup \{\eta\}) > 0,$$
and the flirting of $(p_n)$ with $q_k$ yields the existence of an $m$ with $0 < d(p_m,q_k) < \lambda \leqslant \eta$. This $m$ is $> n$ by construction.
A: This is easier than you think.
Since $p_n$ flirts with each $q_n$ then you have that $d(p_n,q_n)\to 0$. 
Because you are given $q_n\to q$, we have $||q_n-q||$ converges to 0. By the triangle inequality you obtain that 
$d(p_n,q)\le d(p_n,q_n)+  ||q_n-q||$ therefore $d(p_n,q)<\epsilon$ right away. Hence the sequence $p_n$ actually converges to $q$.
Finally the fact $d(p_n,q)<d(p_m,q)$ comes from the fact that the sequence converges hence the distance is getting smaller and smaller, therefore
We can esasily find an $n,m$ with $d(p_n,q)<d(p_m,q)$ by arguing that as $n$ approaches to $\infty$, by the Archimedean Principle, such $n$ exists.
This was easier than we thought!  I love this stuff!
